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Florian Ion PETRESCU & Relly Victoria PETRESCU CAMSHAFT PRECISION COLOR Germany 2013
Citation preview
Florian Ion PETRESCU & Relly Victoria PETRESCU
CAMSHAFT PRECISION
COLOR
Germany 2013
2
Scientific reviewer:
Dr. Veturia CHIROIU
Honorific member of Technical Sciences Academy of Romania (ASTR) PhD supervisor in Mechanical Engineering
Copyright
Title book: Camshaft Precision Color
Authors book: Florian Ion Petrescu & Relly Victoria Petrescu
© 2001-2013, Florian Ion PETRESCU
ALL RIGHTS RESERVED. This book contains material protected under International and Federal Copyright Laws and Treaties. Any unauthorized reprint or use of this material is prohibited. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system without express written permission from the authors / publisher.
Manufactured and published by: Books on Demand GmbH, Norderstedt
ISBN 978-3-8482-2487-6
3
WELCOME
0A
A
B
A-
Fn, vn
Fm, vm
Fa, va
Fi, viFn, vn
Fu, v2
B
B0
A0
A
O
x
e
s0
rb
r0
rA
rB
s
n
C
rb
4
0
A
A
2
B
Fn, vn
Fm, vmFa, va
Fc, vc
Fn, vn
Fu, vu
B
B0
A0
x
rb
r0
rA
rB
A
B
OD
0
d
b
b
r0
G
B
O D
d
A
A0
B0
H
I
l
b
G0
l.’
.’r
M
m
x
1
2
A
Fn; vn
Fm; vm
Fa; va
You are welcome to read the full book! The authors.
5
CONTENT
Welcome............................................................................ 003
Content............................................................................... 005
Cap 01 CAM GEARS EFFICIENCY…………………….… 006
Cap 02 CONTRIBUTIONS AT THE
DYNAMIC OF CAMS ……………………………….……... 013
Cap 03 CAM GEARS DYNAMICS ILLUSTRATED IN THE CLASSIC DISTRIBUTION ……..…………….….. 020
Cap 04 CAM GEARS DYNAMICS
TO THE MODULE B (WITH TRANSLATED
FOLLOWER WITH ROLL)..……………………………….. 030
Cap 05 DYNAMICS OF THE
CLASSIC DISTRIBUTION………………..……………….. 040
Cap 06 PRECISION OF THE
CLASSIC DISTRIBUTION……….……………………….. 046
Cap 07 DYNAMIC SYNTHESIS OF THE ROTARY
CAM AND TRANSLATED TAPPET WITH ROLL..…….. 056
Bibliography........................................................................069
Annex…………………………………………………………..075
6
CHAPTER I
CAM GEARS EFFICIENCY
Abstract: The chapter presents an original method to determine the efficiency of a mechanism with cam and follower. The originality of this method consists in eliminating the friction modulus. In this chapter it analyses four types of cam mechanisms: 1.The mechanism with rotary cam and plate translated follower; 2.The mechanism with rotary cam and translated follower with roll; 3.The mechanism with rotary cam and rocking-follower with roll; 4.The mechanism with rotary cam and plate rocking-follower. For every kind of cam and follower mechanism one uses a different method to determine the most efficient design. We take into account the cam’s mechanism (distribution mechanism), which is the second mechanism in internal-combustion engines. The optimizing of this mechanism (the distribution mechanism), can improve the functionality of the engine and may increase the comfort of the vehicle too.
Keywords: cam, efficiency, translated follower, rocking-follower, follower with roll
1 Introduction
In this chapter the authors present an original method to calculate the efficiency of the cam’s mechanisms. Four kinds of cam and follower mechanisms are analyzed: 1. A mechanism with rotary cam and plate translated follower; 2. A mechanism with rotary cam and translated follower with roll; 3. A mechanism with rotary cam and rocking-follower with roll; 4. A mechanism with rotary cam and plate rocking follower. For every kind of cams and followers mechanism, a different method for the cam’s design with a better efficiency has been utilized.
2 Determining of momentary mechanical efficiency of the rotary cam and plate translated
follower
The consumed motor force, Fc, perpendicular at A to the vector rA, is divided into two components [1, 2]: a) Fm, which represents the useful force, or the motor force reduced to the
follower; b) F, which is the sliding force between the two profiles of cam and follower (Fig. 1). See the written relations (2.1-2.10):
sin cm FF (2.1)
sin12 vv (2.2)
212 sin vFvFP cmu (2.3)
1vFP cc (2.4)
22
1
21 cossin
sin
vF
vF
P
P
c
c
c
ui
(2.5)
22
0
2
2
22
')(
''sin
ssr
s
r
s
A (2.6)
cos cFF (2.7)
7
cos112 vv (2.8)
2
112 cos vFvFP c (2.9)
22
1
21 sincos
cos
vF
vF
P
P
c
c
ci
(2.10)
O
A
r0
s
s’
rA
1vr
2vr
12vr
B
C
D
w
Fr mF
rcFr F
E
© 2002 Florian PETRESCU
The Copyright-Law
Of March, 01, 1989
U.S. Copyright Office
Library of Congress
Washington, DC 20559-6000
202-707-3000
Fig. 1 Forces and speeds to the cam with plate translated follower
3 Determining of momentary dynamic efficiency of the rotary cam and translated follower
with roll
The pressure angle (Fig. 2), is determined by relations (3.5-3.6) [1, 2]. We can write the next forces, speeds and powers (3.13-3.18). Fm, vm, are perpendicular to the vector rA at A. Fm is divided into Fa (the sliding force) and Fn (the normal force). Fn is divided too, into Fi (the bending force) and Fu (the useful force). The momentary dynamic efficiency can be obtained from relation (3.18):
The written relations are the following.
20
22B s)(ser (3.1)
2BB rr (3.2)
B
Br
e sincos (3.3)
8
B
Br
ss 0cossin (3.4)
0A
A
B
A-
Fn, vn
Fm, vm
Fa, va
Fi, viFn, vn
Fu, v2
B
B0
A0
A
O
x
e
s0
rb
r0
rA
rB
s
n
C
rb
Fig. 2 Forces and speeds to the cam with translated follower with roll
22
0
0
)'()(
cos
esss
ss
(3.5)
22
0 )'()(
'sin
esss
es
(3.6)
sinsincoscos)cos( (3.7)
)cos(2222 BbbBA rrrrr (3.8)
9
22
0
220
)'()(
)'()'()(cos
esssr
esressse
A
bA
(3.9)
22
0
2200
)'()(
])'()([)(sin
esssr
resssss
A
bA
(3.10)
cos'
)'()(
')()cos(
220
0
AA
Ar
s
esssr
sss (3.11)
2cos'
cos)cos( A
Ar
s (3.12)
)sin(
)sin(
Ama
Ama
FF
vv (3.13)
)cos(
)cos(
Amn
Amn
FF
vv (3.14)
sin
sin
ni
ni
FF
vv (3.15)
cos)cos(cos
cos)cos(cos2
Amnu
Amn
FFF
vvv (3.16)
mmc
Ammuu
vFP
vFvFP 222 cos)(cos
(3.17)
4
2
2222
22
cos'
]cos'
[]cos)[cos(
cos)(cos
AAA
mm
Amm
c
ui
r
s
r
s
vF
vF
P
P
(3.18)
4 Determining of momentary dynamic efficiency of the rotary cam and rocking follower with
roll
Fm, vm, are perpendicular to the vector rA at A. Fm is divided into Fa (the sliding force) and Fn (the normal force). Fn is divided too into Fc (the compressed force) and Fu (the useful force). The written relations are the following [1, 2] (4.1-4.31).
db
rrdb b
2
)(cos
20
22
0 (4.1)
02 (4.2)
2222 cos)'1(2)'1( bdbdRAD (4.3)
10
0
A
A
2
B
Fn, vn
Fm, vmFa, va
Fc, vc
Fn, vn
Fu, vu
B
B0
A0
x
rb
r0
rA
rB
A
B
OD
0
d
b
b
Fig. 3 Forces and speeds for the rotary cam and rocking follower with roll
RAD
bbd
'cossin 2
(4.4)
RAD
d 2sincos
(4.5)
2222 cos2 dbdbrB (4.6)
B
BB
rd
brd
2cos
222
(4.7)
B
Br
b 2sinsin
(4.8)
cossincossin)sin( 222 (4.9)
sinsincoscos)cos( 222 (4.10)
2
2
BB (4.11)
)sin(cos 2 BB (4.12)
)cos(sin 2 BB (4.13)
)cos(sincos)sin(cos 22 BBB (4.14)
11
)cos(cossin)sin(sin 22 BBB (4.15)
Brrrrr BbbBA cos2222 (4.16) BA
bBA
rr
rrr
2cos
222
(4.17)
Br
r
A
b sinsin (4.18) BA (4.19)
sinsincoscoscos BBA (4.20)
sincoscossinsin BBA (4.21)
2A (4.22) AA
A
cos)cos(sin)sin(
)cos(cos
22
2
(4.23)
cos'
cos
Ar
b (4.24)
2cos
'coscos
Ar
b (4.25)
sin
sin
ma
ma
vv
FF (4.26)
cos
cos
mn
mn
vv
FF (4.27)
sin
sin
nc
nc
vv
FF (4.28)
coscoscos
coscoscos
2 mn
mnu
vvv
FFF (4.29)
mmc
mmuu
vFP
vFvFP 222 coscos
(4.30)
4
2
2222
222
cos'
)cos'
(
)cos(coscoscos
AA
c
ui
r
b
r
b
P
P
(4.31)
5 Determining of momentary mechanical efficiency of the rotary cam and general plate
rocking follower
The written relations are following, (5.1-5.6) (see Fig. 4) [1, 2]:
'1
']sin)(cos)([ 0
20
2
brbrdAH (5.1)
sin)(cos)( 20
20 brdbrbOH (5.2)
222 OHAHr (5.3)
22
2
2
22sin;sin
OHAH
AH
r
AH
r
AH
(5.4)
sincos
;sincos
mmn
mmn
vvv
FFF
(5.5)
12
22
22
2
sinsin
OHAH
AH
vF
vF
vF
vF
P
P
mm
mm
mm
nn
c
ni
(5.6)
r0
G
B
O D
d
A
A0
B0
H
I
l
b
G0
l.’
.’r
M
m
x
1
2
A
Fn; vn
Fm; vm
Fa; va
Fig. 4 Forces and speeds for the rotary cam and general plate rocking follower
6 Conclusions
The follower with roll makes the input-force be divided into several components. This is the reason why, the dynamics and the precise-kinematics (the dynamic-kinematics) of mechanism with rotary cam and follower with roll, are more different and difficult. The presented dynamic efficiency of followers with roll is not the same like the classical mechanical efficiency. For plate followers the dynamic and the mechanical efficiency are the same. This is the great advantage of plate followers.
References
[1] PETRESCU F.I., PETRESCU R.V., Determining the dynamic efficiency of cams. SYROM 2005, Bucharest, Romania, Vol. I, pp. 129-134, 2005. [2] PETRESCU F.I., PETRESCU R.V., POPESCU N., The efficiency of cams. In the Second International Conference “Mechanics and Machine Elements”, Technical University of Sofia, November 4-6, Sofia, Bulgaria, Vol. II, pp. 237-243, 2005.
13
CHAPTER II
CONTRIBUTIONS AT THE
DYNAMIC OF CAMS
ABSTRACT: The chapter presents an original method in determining a general, dynamic and differential equation for the motion of machines and mechanisms, particularized for the mechanisms with rotation cams and followers. This equation can be directly integrated by an original method presented in this chapter. After integration the resulted mother equation may be solved immediately. It presents an original dynamic model with one degree of freedom, with variable internal amortization. It determines the resistant force reduced at the valve (4), the motor force reduced at the valve (5), and the coefficient of variable internal amortization (6). The reduced mass can be calculated with the form (8). The differential motion equation takes the exact form (31), and the approximate form (32). The equation (31) is preparing for its integration with the form (35, 36, 37). The (37) form can be directly integrated and it obtains the parental equation (38). The equation (38) can be arranged in forms (39, 40, 41). The mother equation (41) can be solved directly (42-45), or more elegant with finished differences (48 and 49-50).
Keywords: Motor-force, resistant-force, variable internal amortization, differential equation, valve rocker, valve push rod, valve lifter, valve spring.
1. INTRODUCTION
The chapter presents shortly an original method in determining a general dynamic differential equation, particularized for the mechanisms with rotation cams and followers [1, 2, 3].
This equation can be integrated directly by an original method presented in this chapter.
2. PRESENTING A DYNAMIC MODEL, WITH ONE GRADE
OF FREEDOM, WITH VARIABLE INTERNAL AMORTIZATION
2.1. Determining the amortization coefficient of the mechanism
Starting with the kinematical schema of the classical valve gear mechanism (see the figure 1), one creates the translating dynamic model, with a single degree of freedom (with a single mass), with variable internal amortization (see the picture 2), having the motion equation (1).
The formula (1) is just a Newton equation, where the sum of forces on a single element is 0, [1, 2, 3]:
0)( FxcxkxyKxM (1)
14
Where:
M –the mass of the mechanism, reduced at the valve;
K –the elastically constant of the system;
k –the elastically constant of the valve spring;
c –the coefficient of the system’s amortization;
F0 –the elastically force which compressing the valve spring;
x –the effective displacement of the valve;
ys –the theoretical displacement of the tappet reduced at the valve, imposed by the cam’s profile.
5 w1
2
3
4
A
B
C
D
C0
O
FIG. 1. The kinematical schema of the classical valve gear mechanism
M M
k
kx F
F(t) c .
cx
xx(t)
K(y-x)K
y(t)
w
camã
Fig. 2. Dynamic model with a single liberty, with variable internal amortization
15
The Newton equation (1) can be written in form (2):
)()( 0 xkFxyKxcxM (2)
The differential equation, Lagrange, can be written in form (3).
rm FFxdt
dMxM
2
1 (3)
Comparing the two equations, (2 and 3), we identifie the coefficients and obtain the resistant force (4), the motor force (5) and the coefficient of internal amortization (6), [1, 3]. It can see that the internal amortization coefficient, c, is a variable:
)( 000 xxkxkxkxkFFr (4)
)()( xsKxyKFm (5)
dt
dMc
2
1 (6)
It places the variable coefficient, c, (see the relation 6), in the Newton equation (form 1 or 2) and obtains the equation (7), [1, 3]:
0)(2
1FyKxkKx
dt
dMxM (7)
The reduced mass can be written in form (8), (the reduced mass of the system, reduced at the valve), [1]:
244
211
22325 )()()()(
xJ
xJ
x
ymmmM
ww (8)
With the following notations:
m2 =the mass of the tappet (of the valve lifter);
m3 =the mass of the valve push rod;
m5 =the valve mass;
J1 =the inertia mechanical moment of the cam;
J4 =the inertia mechanical moment of the valve rocker;
2y =the tappet velocity, or the second movement-low, imposed by the cam’s profile;
x =the real (dynamic) valve velocity.
If one notes with i=i25, the ratio of transmission tappet-valve, given from the valve rocker, the theoretically velocity of the valve, y , (the tappet velocity reduced at the valve), takes the form
(9), where the ratio of transmission, i, is given from the formula (10).
i
yyy 2
5
(9)
DC
CCi
0
0 (10)
It can write the following relations (11-16), where y’ is the reduced velocity forced at the tappet by the cam’s profile. With the relations (10, 13, 14, 16) the reduced mass (8), can be written in the forms (17–19):
'1 xx w (11)
''21 xx w (12)
16
'1'212 yiyy ww (13)
'
1
'1
11
xxx
w
ww
(14)
DC
y
DC
CC
CC
y
CC
iy
CC
y
CC
y
0
1
0
0
0
1
0
1
0
'21
0
24
'''.
wwwww
(15)
'
'1
'
'
010
14
x
y
DCxDC
y
x
w
ww
(16)
2
04
21
2325 )
'
'1()
'
1()
'
'()(
x
y
DCJ
xJ
x
yimmmM
(17)
21
2
20
432
25 )
'
1()
'
'(]
)()([
xJ
x
y
DC
JmmimM (18)
21
25 )
'
1()
'
'(*
xJ
x
ymmM (19)
It derivates dM/d and obtains the relations (20–22):
)'
''
'
''()
'
'(2)
'
'''''(
'
'2
'
)''''''(
'
'2])
'
'[(
2
2
2
2
x
x
y
y
x
y
x
yxy
x
y
x
yxxy
x
y
d
x
yd
(20)
32
2
'
''2
'
''
'
2])
'
1[(
x
x
x
x
xd
xd
(21)
312
'
''2)
'
''
'
''()
'
'(*2
x
xJ
x
x
y
y
x
ym
d
dM
(22)
The relation (6) can be written in form (23) and with relation (22), it’s taking the forms (24–25):
w
d
dMc
2 (23)
}'
'')
'
''
'
''()
'
'(]
)()({[
312
20
432
2
x
xJ
x
x
y
y
x
y
DC
Jmmic w (24)
]'
'')
'
''
'
''()
'
'(*[
312
x
xJ
x
x
y
y
x
ymc w (25)
With the notation (26):
20
432
2
)()(*
DC
Jmmim (26)
2.2. Determining the movement equations
With the relations (19, 12, 25, 11) the equation (2) takes the forms (27, 28, 29, 30 and 31):
02 )(''' FyKxkKxcxM ww (27)
17
03122
2221
225
2
)('
''')
'
''
'
''()
'
'(
*''')'
1('')
'
'(*''
FyKxkKx
xJx
x
x
y
y
x
y
mxxx
Jxx
ymmx
w
wwww
(28)
0
2
2222
5
2
)('
'''*
'')'
'(*)
'
'(''*''
FyKxkKx
yym
xx
ym
x
yxmxm
w
www
(29)
02
52
'
'''*)('' FyK
x
yymxkKxm ww (30)
052 )()
'
'''*''( FyKxkK
x
yymxm w (31)
The exact equation (31) can be approximated at the form (32) with x’y’:
052 )()''*''( FyKxkKymxm w (32)
With the following notations: y=s, y’=s’, y’’=s’’, y’’’=s’’’, the equation (32) takes the approximate form (33) and the complete equation (31) takes the exact form (34).
052 )()''*''( FsKxkKsmxm w (33)
052 )()
'
'''*''( FsKxkK
x
ssmxm w (34)
3. SOLVING THE DIFFERENTIAL EQUATION BY DIRECT
INTEGRATION AND OBTAINING THE MOTHER EQUATION
It integrates the equation (31) directly. It prepares the equation (31) for the integration. First, we write (31) in form (35):
I
IIITII
Sx
yymxmxkyKxkK
2*2*
0)(w
w (35)
The equation (35), can be amplified by x’ and obtains the relation (36):
III
T
III
S
III
yymxxm
xxkxyKxxkK
2*2*
0)(
ww (36)
Now, it replaces the term K.y.x’ with IykK
KyK
, (taken in calculation the statically
assumption, Fm=Fr) and it obtains the form (37):
III
T
III
S
III
yymxxm
xxkyykK
KxxkK
2*2*
0
2
)(
ww
(37)
It integrates directly the equation (37) and obtains the mother equation (38):
18
Cy
mx
m
xxky
kK
KxkK
TS
2
'
2
'
22)(
22*
22*
0
222
ww
(38)
With the initial condition, at the =0, y=y’=0 and x=x’=0, it obtains for the constant of integration, C the value 0. In this case the equation (38), takes the form (39):
2
'
2
'
22)(
22*
22*
0
222 ym
xmxxk
y
kK
KxkK TS
ww (39)
The equation (39) can be put in the form (40), if one divides it with the 2
kK :
0)(
''2 2
2
22
2*2
2*02
y
kK
Ky
kK
mx
kK
mx
kK
xkx TS ww
(40)
The mother equation (40), take the form (41), if one notes: '' ykK
Kx
, (the static
assumption, Fm=Fr).
0')(
)(
)(2 22
**
2
2
2
2
202
y
kK
mmkK
K
ykK
Kx
kK
xkx
TS
w (41)
3.1. Solving the mother equation (41) directly
The equation (41) is a two degree equation in x; One determines directly, (42-43) and X1,2 (44):
22
*
2
2*
2
220 '
)(
)(
)(
)()(w
y
kK
mkK
Km
kK
sKxkTS
(42)
22
*
2
2*
2
220 )'(
)(
)(
)(
)()(w
sD
kK
mkK
Km
kK
sKxkTS
(43)
kK
xkX 0
2,1 (44)
Physically, just the positive solution is valid (see the relation 45):
kK
xkX
0 (45)
3.2. Solving the mother equation (41) with finished differences
We can solve the mother equation (41) using the finished differences. We notes:
XsX (46)
19
With the notation (46) placed in the mother equation (41), it obtains the equation (47):
0')(
)(
)(
222)(
22
**
2
2
2
2
2
0022
ykK
mmkK
K
skK
K
XkK
xks
kK
xksXXs
TS
w
(47)
The equation (47) is a two degree equation in X, which can be solved directly with (49)
and X1,2, (50), or transformed in a single degree equation in X, with (X)20, solved by the
relation (48).
20
22**2
0
22
)()(2
)'(])([)(2)2(
)1(
kKkK
xks
DsmkKmkK
KskKxksKkk
XTS
w (48)
2
22**2
20
222
)(
)'(])([
kK
sDmkKmkK
KxksK TS
w
(49)
)( 0
kK
xksX
(50)
CONCLUSION
The direct integration of the differential equation (31) generates the mother equation (41), which can be solved directly, with the relation (48). “D” represents the dynamic transmission function (the dynamic transmission coefficient).
REFERENCES
[1] Antonescu, P., Oprean, M., Petrescu, Fl., Analiza dinamică a mecanismelor de distribuţie cu came, In: The Proceedings of 7
th National Symposium on RIMS, MERO’87, Bucureşti, vol. 3, pp.
126-133, 1987. [2] Antonescu, P., Petrescu, Fl., Contributii la analiza cinetoelastodinamică a mecanismelor de distribuţie, In: The Proceedings of 5
th International Symposium on TMM, SYROM’89, Bucureşti, pp.
33-40, 1989. [3] Petrescu, F., Petrescu, R., Elemente de dinamica mecanismelor cu came, In: The Proceedings of 7
th National Symposium, PRASIC’02, Braşov, vol. I, pp. 327-332, 2002.
20
CHAPTER III CAM GEARS DYNAMICS ILLUSTRATED
IN THE CLASSIC DISTRIBUTION
Abstract: The chapter presents an original method to determine the general dynamics of mechanisms with rotation cams and followers, particularized to the plate translated follower. First, it presents the dynamics kinematics. Then it solves the Lagrange equation and using an original dynamic model with one degree of freedom, with variable internal amortization, it makes the dynamic analysis. Keywords: cam dynamics, classic distribution, cams, followers, dynamics
1 Introduction
The chapter proposes an original dynamic model illustrated for the rotating cam with plate translate follower. It presents the dynamics kinematics (the original kinematics); the variable velocity of the camshaft obtained by an approximate method is used with an original dynamic system having one degree of freedom and a variable internal amortization [1]; it tests two movement laws, one classic and the other original.
2 Dynamics of the classic distribution mechanism 2.1 Precision kinematics in the classic distribution mechanism
In the picture number one, it presents the kinematic schema of the classic distribution mechanism, in two consecutive positions; with an interrupted line is represented the particular position when the follower is situated in the lowest possible plane, (s=0), and the cam which has a
clockwise rotation, with constant angular velocity, w, is situated in the point A0, (the fillet point between the base profile and the rise profile), a particular point that marks the beginning of the rise movement of the follower, imposed by the cam-profile; with a continue line is represented the higher joint in a certain position of the rise phase.
O
Ai
r0=s0
s
s’
rA
A0
1vr
2vr
12vr
B
C
D
A0i
w
Fig. 1 The kinematics of the classic distribution mechanism
21
The point A0, which marks the initial higher pair, represents in the same time the contact point between the cam and the follower in the first position. The cam is rotating with the angular
velocity, w (the camshaft angular velocity), describing the angle , which shows how the base circle has rotated clockwise (together with the camshaft); this rotation can be seen on the base circle between the two particular points, A0 and A0i.
In this time the vector rA=OA (which represents the distance between the centre of cam O,
and the contact point A), has rotated anticlockwise with the angle . If one measures the angle , which positions the general vector, rA, in function of the particular vector, rA0, it obtains the relation (0):
(0)
where rA is the module of the vector Arr
, and A represents the phase angle of the vector Arr
.
The angular velocity of the vector Arr
is A which is a function of the angular velocity of the
camshaft, w, and of the angle (by the movement laws s(), s’(), s’’()).
The follower isn’t acted directly by the angle and the angular velocity w; it’s acted by the
vector Arr
, which has the module rA, the position angle A and the angular velocity A . From here
we deduce a particular (dynamic) kinematics, the classical kinematics being just static and approximate kinematics.
Kinematic, it defines the next velocities (Fig. 1).
1vr
=the cam’s velocity; which is the velocity of the vector Arr
, in the point A; now the classical
relation (1) becomes an approximate relation, and the real relation takes the form (2).
w.1 Arv (1)
AArv .1 (2)
The velocity ACv 1
r is separating into the velocity 2v
r=BC (the follower’s velocity which
acts on its axe, vertically) and 12vr
=AB (the slide velocity between the two profiles, the sliding
velocity between the cam and the follower, which works along the direction of the commune tangent line of the two profiles in the contact point).
Because usually the cam profile is synthesis for the classical module C with the AD=s’ known, we can write the relations:
22
0
2 ')( ssrrA (3)
22
0 ')( ssrrA (4)
22
0
00
')(cos
ssr
sr
r
sr
A
(5)
22
0 ')(
''sin
ssr
s
r
s
r
AD
AA (6)
A
A
AA sr
srvv '.
'..sin.12 (7)
Now, the follower’s velocity isn’t s ( w '2 ssv ), but it’s given by the relation (9). In the
case of the classical distribution mechanism the transmitting function D is given by the relations (8):
22
w
w
A
A
D
D
.
(8)
w .'.'.2 Dssv A (9)
The determining of the sliding velocity between the profiles is made with the relation (10):
A
A
AA srr
srrvv ).(..cos. 0
0
112
(10)
The angles and A will be determined, and also their first and second derivatives.
The angle has been determined from the triangle ODAi (Fig.1) with the relations (11-13):
22
0 ')(
'sin
ssr
s
(11)
22
0
0
')(cos
ssr
sr
(12)
sr
stg
0
' (13)
It derives (11) in function of angle and obtains (14):
22
0
0
')(
'''.').('.'.'
cos'.ssr
r
ssssrsrs
A
A
(14)
The relation (14) will be written in the form (15):
22
0
22
0
2
0
222
0
')(].')[(
''.').('''.')'.('cos'.
ssrssr
sssrssssrs
(15)
From the relation (12) it extracts the value of cos, which will be introduced in the left term of the expression (15); then we can reduce s’’.s’2 from the right term of the expression (15) and it obtains the relation (16):
22
0
22
0
2
00
22
0
0
')(].')[(
]')'.(').[(
')('.
ssrssr
ssrssr
ssr
sr
(16)
After some simplifications the relation (17), which represents the expression of ’, is finally obtained:
22
0
2
0
')(
')'.(''
ssr
ssrs
(17)
Now when ’ has been explicitly deduced, the next derivatives can be determined. The expression (17) will be derived directly and it obtains for the beginning the relation (18):
23
222
0
0
2
0
22
00
]')[(
]'''')][(')(''[2]')][('''2''')('''[''
ssr
ssssrssrsssrsssssrs
(18)
The terms from the first bracket of the numerator (s’.s’’) are reduced, and then it draws out s’ from the fourth bracket of the numerator and obtains the expression (19):
222
0
0
2
0
22
00
]')[(
]''].[')'.(''.[.2]')].[('''.)'.(''[''
ssr
ssrssrssssrsssrs
(19)
Now we can calculate A, with its first two derivatives, A and A
. We will write instead of
A, to simplify the notation. It determines the relation (20) which is the same of (0):
(20)
We derive the relation (20) and one obtains the expression (21):
wwww .)'1.('. D (21)
It derives twice (20), or derives (21) and obtains (22):
22 ''' ww D (22)
We can write now the transmission functions, D and D’ (for the classical module, C), in the forms (23-24):
1'D (23)
''ID (24)
To calculate the follower’s velocity (25) we need the expression of the transmission function, D.
DsDssswsv A w ''''2 (25)
Where:
w.Dw (26)
For the classical distribution mechanism (Module C), the variable w is the same as A (see
the relation 25).
But in the case of B and F modules (at the cam gears where the follower has a roll), the transmitted function D and w take complex forms.
We can determine now the acceleration of the follower (27).
2
2 )''''( w DsDsay (27)
Figure 2 represents the classical and dynamic kinematics; the velocities (a), and the accelerations (b).
24
-4
-3
-2
-1
0
1
2
3
4
0 50 100 150 200
Vclasic[m/s]
Vprecis[m/s]
Fig. 2a The classical and dynamic kinematics; velocities of the follower
-4000
-3000
-2000
-1000
0
1000
2000
3000
4000
5000
0 50 100 150 200
a2clasic[m/s2]
a2precis[m/s2]
Fig. 2b The classical and dynamic kinematics; accelerations of the follower
25
To determine the acceleration of the follower, s’ and s’’, D and D’, ’ and ’’ are necessary be known.
The dynamic kinematics diagrams of v2 (obtained with relation 25, see Fig. 2a), and a2 (obtained with relation 27, see Fig. 2b), have a more dynamic aspect than one kinematic (classic).
It has used the movement law SIN, a rotational speed of the crankshaft n=5500 rpm, a rise
angle u=750, a fall angle d=750 (identically with the ascendant angle), a ray of the basic circle of the cam, r0=17 mm and a maxim stroke of the follower, hT=6 mm.
Anyway, the dynamics is more complex, having in view the masses and the inertia moments, the resistant and motor forces, the elasticity constants and the amortization coefficient of the kinematic chain, the inertia forces of the system, the angular velocity of the camshaft and
the variation of the camshaft’s angular velocity, w, with the cam’s position, , and with the rotational speed of the crankshaft, n.
2.2 Solving approximately the Lagrange movement equation
In the kinematics and the static forces study of the mechanisms one considers the shaft’s
angular velocity constant, w =constant, and the angular acceleration null, 0 w . In
reality, this angular velocity w isn’t constant, it is variable with the camshaft position, .
In mechanisms with cam and follower the camshaft’s angular velocity is variable as well.We shall see further the Lagrange equation, written in the differentiate mode and its general solution.The differentiate Lagrange equation has the form (28):
*2** ..
2
1. MJJ I (28)
Where J* is the mechanical inertia moment (mass moment, or mechanic moment) of the mechanism, reduced at the crank, and M* represents the difference between the motor moment
reduced at the crank and the resistant moment reduced at the crank; the angle represents the rotation angle of the crank (crankshaft). J*I represents the derivative of the mechanic moment in
function of the rotation angle of the crank (29).
Ld
dJJ I
** .
2
1.
2
1 (29)
Using the notation (29), the equation (28) will be written in the form (30):
*2* .. MLJ (30)
We divide the terms by J* and (30) takes the form (31):
*
*2
*.
J
M
J
L (31)
The term with 2 will be moved to the right side of the equation and the form (32) will be
obtained:
2
**
*
. J
L
J
M (32)
26
Replacing the left term of the expression (32) with (33) we obtain the relation (34):
w
w
...
d
d
d
d
dt
d
d
d
dt
d
(33)
*
2*2
**
* ...
J
LM
J
L
J
M
d
d ww
ww
(34)
Because, for an angle , w is different from the nominal constant value wn, it can write the
relation (35), where dw represents the momentary variation for the angle ; the variable dw and
the constant wn lead us to the needed variable, w:
www dn (35)
In the relation (35), w and dw are functions of the angle , and wn is a constant parameter, which can take different values in function of the rotational speed of the drive-shaft, n. At a
moment, n is a constant and wn is a constant as well (because wn is a function of n). The angular
velocity, w, becomes a function of n too (see the relation 36):
))(,()(),( ndnn nn wwww (36)
With (35) in (34), it obtains the equation (37):
wwwww ddJ
L
J
Mdd nn ].).([).( 2
**
*
(37)
The relation (37) takes the form (38):
]..2)(.[..)(. 22
**
*2 wwwwwww ddd
J
Ld
J
Mdd nnn (38)
The equation (38) will be written in the form (39):
0....2).(.
...)(.
*
2
*
2
**
*2
www
wwww
ddJ
Ldd
J
L
dJ
Ld
J
Mdd
n
nn
(39)
The relation (39) takes the form (40):
0)...(
.).2
1..(2)).(1.(
2
**
*
*
2
*
n
n
dJ
Ld
J
M
ddJ
Ldd
J
L
w
www
(40)
The relation (40) is an equation of the second degree in dw. The discriminate of the equation (40) can be written in the forms (41) and (42):
2
*
22
2*
2
*
*
2
2*
*2
*
222
2*
2
...).(.
).(.
..4
.).(
nn
nn
n
dJ
Ld
J
Ld
J
M
dJ
MLd
J
Ld
J
L
ww
ww
w
(41)
w
dJ
Md
J
MLn .).(.
4 *
*2
2*
*2
(42)
27
We keep for dw just the positive solution, which can generate positives and negatives
normal values (43), and in this mode only normal values will be obtained for w; for 0 it
considers dw=0 (this case must be not seeing if the equation is correct).
1.
2..
*
*
ww
w
dJ
L
dJ
L
d
n
n
(43)
Observations: For mechanisms with rotate cam and follower, using the new relations, with M* (the reduced moment of the mechanism) obtained by the writing of the known reduced resistant moment and by the determination of the reduced motor moment by the integration of the resistant
moment it frequently obtains some bigger values for dw, or zones with negative, with complex
solutions for dw. This fact gives us the obligation to reconsider the method to determine the reduced moment.
If we take into consideration M*r and M*m, calculated independently (without integration), it
obtains for the mechanisms with cam and follower normal values for dw, and 0 .
In paper [1] it presents the relations to determine the resistant force (44) reduced to the valve, and the motor force (45) reduced to the ax of the valve:
).( 0
* xxkFr (44)
).(* xyKFm (45)
The reduced resistant moment (46), or the reduced motor moment (47), can be obtained by the resistant or motor force multiplied by the reduced velocity, x’.
')..( 0
* xxxkM r (46)
')..(* xxyKM m (47)
2.3 The dynamic relations used
The dynamics relations used (48-49), have been deduced in the paper [1]:
20
22**2
022
)()(2
)'(])([)(2)2(
)1(
kKkK
xks
DsmkKmkK
KskKxksKkk
XTS
w (48)
20
0
22
20
22**2
)()(2
)(2)2(
)()(2
)'(])([
kKkK
xks
skKxksKkk
kKkK
xks
DsmkKmkK
K
sXTS
w
(49)
2.4 The dynamic analysis
The dynamic analysis or the classical movement law sin, can be seen in the diagram from figure 3, and in figure 4 one can see the diagram of an original movement law (C4P) (module C).
28
-2000
-1000
0
1000
2000
3000
4000
5000
6000
0 50 100 150 200
a[m/s2]
673.05s*k[mm] k=
n=5000[rot/min]
u=75 [grad]
k=20 [N/mm]
r0=14 [mm]
x0=40 [mm]
hs=6 [mm]
hT=6 [mm]
i=1;=8.9%
legea: sin-0
y=x-sin(2x)/(2)
Analiza dinamicã la cama rotativã cu tachet
translant plat - A10amax=4900
s max =5.78
amin= -1400
Fig. 3 The dynamic analysis of the law sin, Module C, u=750,
n=5000 rpm
-10000
0
10000
20000
30000
40000
50000
0 20 40 60 80 100a[m/s2]
7531.65s*k[mm] k=
n=10000[rot/min]
u=45 [grad]
k=200 [N/mm]
r0=17 [mm]
x0=50 [mm]
hs=6 [mm]
hT=6 [mm]
i=1;=15.7%
legea:C4P1-1
y=2x-x2
yc=1-x2
Analiza dinamicã la cama rotativã cu tachet
translant plat - A10
amax=39000
s max =4.10
amin= -8000
Fig. 4 The dynamic analysis of the new law, C4P, Module C, u=450,
n=10000 rpm
29
-40000
-20000
0
20000
40000
60000
80000
100000
120000
0 50 100 150 200
a[m/s2]
19963,94s*k[mm] k=
n=40000[rot/min]
u=80 [grad]
k=400 [N/mm]
r0=13 [mm]
x0=150 [mm]
hs=10 [mm]
hT=10 [mm]
i=1;=12.7%
rb=2 [mm]
e=0 [mm]
legea: C4P1-5
y=2x-x2
Analiza dinamicã la cama rotativã cu tachet
translant cu rolã
amax=97000
smax=3.88
amin= -33000
Fig. 5 Law C4P1-5, Module B, u=800, n=40000 rpm
-60000
-40000
-20000
0
20000
40000
60000
80000
100000
0 50 100 150 200
a[m/s2]
15044,81s*k[mm] k=
n=40000[rot/min]
u=85 [grad]
k=800 [N/mm]
r0=10 [mm]
rb=3 [mm]
b=30 [mm]
d=30 [mm]
x0=200 [mm]
i=1;=16.5%
legea: C4P3-2
y=2x-x2
yc=1-x2
Analiza dinamicã la cama rotativã cu tachet
balansier cu rolã (Modul F) - A12amax=80600
s max=4.28
amin= -40600
hT=15.70 [mm]
Fig. 6 Law C4P3-2, Module F, u=850, n=40000 rpm
3 Conclusions
Using the classical movement laws, the dynamics of the distribution cam-gears depreciate rapidly at the increasing of the rotational speed of the shaft. To support a high rotational speed it is necessary the synthesis of the cam-profile by new movement laws, and for the new Modules.
A new and original movement law is presented in the pictures number 4, 5 and 6; it allows the increase of the rotational speed to the values: 10000-20000 rpm, in the classical module C presented (Fig. 4). With others modules (B, F) it can obtain 30000-40000 rpm (see Figs. 5, 6).
References
[1] Petrescu F.I., Petrescu R.V., Contributions at the dynamics of cams. In the Ninth IFToMM International Sympozium on Theory of Machines and Mechanisms, SYROM 2005, Bucharest, Romania, Vol. I, pp. 123-128, 2005.
30
CHAPTER IV
CAM GEARS DYNAMICS TO THE MODULE B (WITH TRANSLATED FOLLOWER WITH ROLL)
Abstract: The chapter briefly presents an original method for determining the dynamics of mechanisms with rotation cam and translated follower with roll. First, one presents the dynamics kinematics. Then one performs the dynamic analysis of a few models, for some movement laws, imposed on the follower, by the designed cam profile. Keywords: cam dynamics, translated follower with roll, movement laws, dynamics kinematic
1 Introduction
The chapter proposes an original dynamic model of the cam gear with a translated follower with a roll. First, one presents the dynamics kinematics. Then one performs the dynamic analysis of a few models, for some movement laws, imposed on the follower, by the designed cam profile.
2 The dynamics of distribution mechanisms with translated follower with roll 2.1 Generalities
The angle 0 defines the basic position of the vector, 0Br , in the OCB0 triangle having a
right angle (1-4):
bB rrr 00 (1)
22
0 0ers B (2)
0
0cosBr
e (3)
0
0
0sinBr
s (4)
The pressure angle, , between the normal n (which passes through the contact point A) and a vertical line, can be calculated with relations (5-7).
22
0
0
)'()(cos
esss
ss
(5)
22
0 )'()(
'sin
esss
es
(6)
ss
estg
0
' (7)
The vector Ar can be determined with relations (8-9):
2
0
22 )cos()sin( bbA rssrer (8)
2
0
2 )cos()sin( bbA rssrer (9)
31
0A
A
B
A-
Fn, vn
Fm, vm
Fa, va
Fi, viFn, vn
Fu, v2
B
B0
A0
A
O
x
e
s0
rb
r0
rA
rB
s
n
C
rb
Fig. 1 Mechanism with rotating cam and translating follower with roll
We can calculate A (10-11):
A
b
Ar
re
sincos
(10)
A
b
Ar
rss
cossin 0
(11)
2.2 The relations to design the profile
0 A (12)
00 sinsincoscoscos AA (13)
00 sincoscossinsin AA (14)
32
A (15)
sinsincoscoscos A (16)
cossincossinsin A (17)
2.3 The exact kinematics of B Module
From the triangle OCB (fig. 1) the length rB (OB) and the complementary angles B (COB)
and (CBO) are determined.
2
0
22 )( sserB (18)
2
BB rr (19)
B
Br
e sincos (20)
B
Br
ss 0cossin (21)
From the general triangle OAB, where one knows OB, AB, and the angle between them, B
(ABO, which is the sum of and ), the length OA and the angle (AOB) can be determined:
sinsincoscos)cos( (22)
)cos(2222 BbbBA rrrrr (23)
BA
bBA
rr
rrr
2cos
222
(24)
cossincossin)sin( (25)
)sin(sin A
b
r
r (26)
With B and we can deduce now A and A :
BA (27)
BA (28)
From (20) one obtains B (32), (see 29-32) where Br (31) can be deduced from (18). Then, (33)
will be obtained from (24):
2
sinB
B
BBr
re
(29)
2
0 )( B
BB
Brss
rre
(30)
sssrrsssrr BBBB )()(22 00 (31)
22
0
0
)(
)(
BB
Br
se
rss
ssse
(32)
33
BBAABA
BABA
rrrrrr
rrrr
22sin2
cos2cos2
(33)
From (33) one writes (38), but it is necessary to obtain first Ar (34) from expression
(23):
)()sin(2
)cos(222
Bb
BbBBAA
rr
rrrrrr (34)
To solve (34) we need the derivatives and . From (7) relations (35 and 36) will be
obtained. takes the form (37):
22
0
0
)'()(
)'(')('''
esss
essess
(35) w ' (36)
2
B
Br
se
(37)
Now we can determine (38), A (28) and A (39):
sin
coscos
BA
BBAABABA
rr
rrrrrrrr (38)
AA w (39)
We write cos A and sin A (40-41):
22
0
22
0
)'()(
)'()'()(cos
esssr
esressse
A
b
A
(40)
22
0
22
00
)'()(
])'()([)(sin
esssr
resssss
A
b
A
(41)
Further, we can obtain expression cos(A-) (42), and cos(A-).cos (43):
cos'
)'()(
')()cos(
22
0
0
AA
Ar
s
esssr
sss (42)
2cos'
cos)cos( A
Ar
s (43)
Finally the forces and the velocities are deduced as follows (48-50):
)sin(
)sin(
Ama
Ama
FF
vv
(44)
)cos(
)cos(
Amn
Amn
FF
vv
(45)
34
sin
sin
ni
ni
FF
vv
(46)
cos)cos(cos
cos)cos(cos2
Amnu
Amn
FFF
vvv
(47)
2.4 Determining the efficiency of the Module B
22
2 cos)(cos Ammuu vFvFP (48)
mmc vFP (49)
4
2
222
222
22
cos'
]cos'
[
]cos)[cos(cos)(cos
cos)(cos
AA
AA
mm
Amm
c
ui
r
s
r
s
vF
vF
P
P
(50)
2.5 Determining the transmission function D, for the Module B
The follower’s velocity (47) can be written into the form (51):
w
222
2
2
cos'cos'cos'
cos'
cos)cos(cos
ssr
sr
r
svvvv
I
AA
A
AA
A
mAmn
(51)
With relations (51) and (52) we determine the transmission function (the dynamic modulus), D (53):
w Dsv '2 (52)
2cos I
AD (53)
Expression cos2 is known (54):
22
0
2
02
)'()(
)(cos
esss
ss
(54)
The expression of the ’A is more difficult (55):
35
]}')[(2)'()(
])/{[(])'()[(
/]})'()()'(')(''[
)'()(])'(){[(
])'()(')[(
22
0
22
0
222
0
22
0
22
00
22
0
22
0
22
0
22
0
seessresss
ressesss
esssesssssr
esssesss
esssrseess
b
b
b
b
I
A
(55)
We will determine by its expressions (56-57):
22
0
22
0
22
0
22
0
)'()(
]')[()'()(])[(cos
esssrr
seessresssess
BA
b
(56)
22
0
0
)'()(
')(sin
esssrr
sssr
BA
b
(57)
2.6 The dynamics of the Module B
For the dynamics of the Module B the relations (58-60) are used:
][2
'
])(
[2
)(
2
0
2
2**
2
2
02
2
2
kK
kxs
ykK
mmkK
K
skK
kxs
kK
kKk
X
TS
w
(58)
][2
)'(
])(
[2
)(
2
0
2
2**
2
2
02
2
2
kK
kxs
sDkK
mmkK
K
skK
kxs
kK
kKk
X
TS
w
(59)
XsX (60)
2.7 The dynamic analysis of the module B
It presents now the dynamics of the module B for some known movement laws.
We begin with the classical law SIN (see the diagram in figure 2); A speed rotation n=5500 [rot/min], for a maxim theoretical displacement of the valve h=6 [mm] is used. The phase angle is
u=c=65 [degree]; the ray of the basic circle is r0=13 [mm].
For the ray of the roll the value rb=13 [mm] has been adopted.
36
-4000
-2000
0
2000
4000
6000
8000
0 50 100 150
a[m/s2]
880.53s*k[mm] k=
n=5500[rot/min]
u=65 [grad]
k=30 [N/mm]
r0=13 [mm]
x0=20 [mm]
hs=6 [mm]
hT=6 [mm]
i=1;=11.5%
rb=13 [mm]
e=6 [mm]
legea: sin-0
y=x-sin(2x)/(2)
Analiza dinamicã la cama rotativã cu tachet
translant cu rolãamax=6400
smax=5.81
amin= -3000
Fig. 2 The dynamic analysis of the module B. The law SIN, n=550 rpm, u=650,
r0=13 [mm], rb=13 [mm], hT=6 [mm], e=0 [mm],k=30 [N/mm], and x0=20 [mm].
-15
-10
-5
0
5
10
15
20
-20 -10 0 10 20
yC [mm]
PROFIL Camã rotativã cu tachet translant cu rolã
u= 65[grad]
c= 65[grad]
r0= 13[mm]
rb = 13[mm]
e= 6[mm]
hT= 6[mm]
Legea SIN
Suportã o turatie n=5500[rot/min]
w
Fig. 3 The profile SIN at the module B. n=5500 rpm
u=650, r0=13 [mm], rb=13 [mm], hT=6 [mm].
37
The dynamics are better than for the classical module C. For a phase angle of just 65 degrees the accelerations have the same values as for the classical module C for a relaxed phase (750-800).
In figure 3 we can see the cam’s profile. It uses the profile sin, a rotation speed n=5500
rpm, and u=650, r0=13 [mm], rb=13 [mm], hT=6 [mm].
The law COS can be seen in figures 4 and 5.
In the figure 4 is presented the dynamic analyze of the profile cos, and its profile design can be seen in the figure 5.
The principal parameters are:
Law COS, n=5500 rpm, u=650, r0=13 [mm], rb=6 [mm], hT=6 [mm], =10.5%.
-3000
-2000
-1000
0
1000
2000
3000
4000
5000
0 50 100 150
a[m/s2]
601.01s*k[mm] k=
n=5500[rot/min]
u=65 [grad]
k=30 [N/mm]
r0=13 [mm]
x0=30 [mm]
hs=6 [mm]
hT=6 [mm]
i=1;=10.5%
rb=6 [mm]
e=0 [mm]
legea: cos-0
y=.5-.5cos(x)
Analiza dinamicã la cama rotativã cu tachet
translant cu rolãamax=4300
smax=5.74
amin= -2000
Fig. 4 The dynamic analysis of the module B. Law COS, n=5500 rpm, u=650, r0=13 [mm], rb=6
[mm], hT=6 [mm], =10.5%.
-15
-10
-5
0
5
10
15
20
-20 -10 0 10 20
yC [mm]
PROFIL Camã rotativã cu tachet translant cu rolã
u= 65[grad]
c= 65[grad]
r0= 13[mm]
rb = 6[mm]
e= 0[mm]
hT= 6[mm]
Legea COS
Suportã o turatie n=5500[rot/min]
w
Fig. 5 The profile COS at the module B, n=5500 rpm, u=650, r0=13 [mm], rb=6 [mm], hT=6 [mm].
38
-2000
0
2000
4000
6000
8000
10000
12000
14000
0 50 100 150 200 a[m/s2]
1896.75s*k[mm] k=
n=5500[rot/min]
u=80 [grad]
k=50 [N/mm]
r0=13 [mm]
x0=50 [mm]
hs=6 [mm]
hT=6 [mm]
i=1;=8.6%
rb=6 [mm]
e=0 [mm]
legea: C4P1-0
y=2x-x2
Analiza dinamicã la cama rotativã cu tachet
translant cu rolãamax=13000
smax=5.37
amin= -600
Fig. 6 The dynamic analyze. Law C4P1-0, n=5500 rpm, u=800, r0=13 [mm], rb=6 [mm], hT=6 [mm].
In figure 6 the law C4P, created by the authors, is analyzed dynamic. The vibrations are diminished, the noises are limited, the effective displacement of the valve is increased, smax=5.37 [mm].
-20
-15
-10
-5
0
5
10
15
20
25
-20 -10 0 10 20
yC [mm]
PROFIL Camã rotativã cu tachet translant cu rolã
u= 80[grad]
c= 80[grad]
r0= 13[mm]
rb = 3[mm]
e= 0[mm]
hT= 6[mm]
Legea C4P1-0
Suportã o turatie n=5500[rot/min]
w
Fig. 7 The profile C4P of the module B.
The efficiency has a good value =8.6%. In figure 7 the profile of C4P law is presented. It starts at the law C4P with n=5500 [rpm], but for this law the rotation velocity can increase to high values of 30000-40000 [rpm] (see Fig. 8).
39
-40000
-20000
0
20000
40000
60000
80000
100000
120000
0 50 100 150 200
a[m/s2]
19371.43s*k[mm] k=
n=40000[rot/min]
u=80 [grad]
k=400 [N/mm]
r0=13 [mm]
x0=150 [mm]
hs=10 [mm]
hT=10 [mm]
i=1;=14.4%
rb=6 [mm]
e=0 [mm]
legea: C4P1-5
y=2x-x2
Analiza dinamicã la cama rotativã cu tachet
translant cu rolã
amax=94000
smax=3.88
amin= -33000
Fig. 8 The dynamic analysis of the module B. Law C4P1-5, n=40000 rpm.
3 Conclusions
We can speak about an advantage of the module B in comparison to the classical module C. With the module B, (when the follower is provided with a roll) it can obtain high rotation velocity with superior efficiency.
References
[1] Petrescu F.I., Petrescu R.V., Contributions at the dynamics of cams. In the Ninth IFToMM International Sympozium on Theory of Machines and Mechanisms, SYROM 2005, Bucharest, Romania, Vol. I, pp. 123-128, 2005.
40
CHAPTER V DYNAMICS OF THE CLASSIC DISTRIBUTION
Abstract: This chapter presents an original methods to determine the dynamic parameters at the camshaft (the distribution mechanisms). We determine initially the mass moment of inertia (mechanical) of the mechanism, reduced to the element of rotation, ie at cam (basically using kinetic energy conservation, the system 1). Average moment of inertia is calculated with equation (2). The expression (2) depends on the type of cam-tappet mechanism, and of the law of motion used both uphill and downhill. The angular velocity
is a function of the position cam () but also of its speed (3). To determine ω2 (relationship 3) have found J *,
and more specifically Jmax. Differentiating the formula (6), against time, is obtained the angular
acceleration expression (8). Differentiating twice successively, the expression (9) in the angle , we obtain a reduced tappet speed (equation 10), and reduced tappet acceleration (11). The real and dynamic, tappet acceleration can be determined directly using the relation (12). General (original) dynamic equations of
motion for the determination of ω and have the form (13).
Keywords: cam, cams, cam mechanisms, distribution mechanisms, camshaft, tappet.
We determine initially the mass moment of inertia (mechanical) of the mechanism, reduced to the element of rotation, ie at cam (basically using kinetic energy conservation, the system 1).
22
0
2
var
tan
*
22
0
22
0
*
222
0
*
22
0
22
0
2
2
''2
1
2
1
''2
1
2
1
2
1
''2
1
'2
1
'
2
1
smsMsRMsMJJ
JJJ
smsMsRMsMRMJ
smssRMJ
ssRMJ
ssRR
RMJ
Tccciabil
tcons
Tcccc
Tc
ccama
ccama
(1)
Average moment of inertia is calculated with equation (2).
22
1
2
max2
0
*
max
*
min* JRM
JJJ cm
(2)
The expression (2) depends on the type of cam-tappet mechanism, and of the law of
motion used both uphill and downhill. The angular velocity is a function of the position cam () but also of its speed (3).
41
*
2*2
J
J mm ww
(3)
O
A
r0
s
s’
rA
1vr
2vr
12vr
B
C
D
w
Fr mF
rcFr F
E
Fig. 1 Forces and speeds to the cam with plate translated follower
To determine ω2 (relationship 3) have found J *, and more specifically Jmax.
And at the classic distribution (rotative cam, and plat tappet in translational motion), the relationship which determine the Jmax, depend and of the law of movement.
We start the simulation with a classical law of motion, namely the cosine law. At the climb, the cosine law is expressed by relations (4).
uu
r
uu
r
uu
r
u
hs
has
hvs
hhs
sin2
'''
cos2
''
sin2
'
cos22
3
3
2
2 (4)
Where varies from 0 to u. It achieves Jmax for =u/2.
42
2
22
2
22
0
2
max48
1
28 u
T
u
c
hm
hhR
hMJ
(5)
The expression (3) now takes the form (6).
B
A
smsMsRMsMRMB
hm
hM
hRMh
MRMA
B
A
m
Tcccc
u
T
u
c
ccc
m
ww
ww
22
0
22
0
2
22
2
22
0
22
0
22
'2'2
4
1
8
1
2
1
8
(6)
Where ωm is the nominal cam velocity and it express at the distribution mechanisms, based on the speed shaft, with relationship (7).
60260
2
6022
nnn motorccm
w (7)
Differentiating the formula (6), against time, is obtained the angular acceleration expression (8).
B
ssmsMRMsM Tccc '''2''02 w (8)
For a classic cam and tappet mechanism (without valve) dynamic movement tappet is expressed by equation (9), who was presented and derived in Chapter 2 (equation 48), and now by canceling valve mass, will customize and reaching form below (9).
kK
xkskK
skKxksKkksmkKsx T
02
0
2222
)(2
)(2)2(')( w (9)
Where x is the dynamic movement of the pusher, while s is its normal, kinematics movement. K is the spring constant of the system, and k is the spring constant of the tappet spring. It note, with x0 the tappet spring preload, with mT the mass of the tappet, with ω the angular
rotation speed of the cam (or camshaft), where s’ is the first derivative in function of of the tappet
movement, s. Differentiating twice successively, the expression (9) in the angle , we obtain a reduced tappet speed (equation 10), and reduced tappet acceleration (11).
43
2
02
0
0
22
0
2222
2
''
'
'2'22'''2
)(2)2(')(
kK
kxskK
Msx
sNkK
kxs
skKkxsskKkssmkKM
skKxksKkksmkKN
T
T
w
w
(10)
3
02
00
0
22
22
0
0
22
0
2222
2
'2''
''''
''2'''22
''''''2
'
'2'22'''2
)(2)2(')(
kK
kxskK
sMkK
kxssN
kK
kxsO
sx
skKxksssKkk
sssmkKO
sNkK
kxs
skKkxsskKkssmkKM
skKxksKkksmkKN
T
T
T
w
w
w
(11)
The real and dynamic, tappet acceleration can be determined directly using the relation (12).
w ''' 2 xxx (12)
General (original) dynamic equations of motion for the determination of ω and have the form (13).
*
*'2
*
*2
*
*2
2
1
;
J
J
J
J
J
Jm
mm
m
w
wwww
(13)
With a program (written in excel) one obtains the diagrams of the movement laws (see the Figure 2), the dynamic tappet acceleration for a n=5500 [rpm] (Fig. 3), and the cam profile (Fig. 4).
44
Fig. 2 s, s’, s’’ diagrams at the cam with plate translated follower
Fig. 3 The dynamic tappet acceleration at the cam with plate translated follower
The profile synthesis was made with the system of relations (14) when the cam is moving in the orar sense, and (15) when the cam is rotating trigonometric.
45
sin'cos)(
sin)(cos'
0
0
ssry
srsx
c
c (14)
sin'cos)(
sin)(cos'
0
0
ssry
srsx
c
c (15)
Fig. 4 The cam profile, at the cam with plate translated follower
46
CHAPTER VI PRECISION OF THE CLASSIC DISTRIBUTION
Abstract: This chapter presents an original methods to determine the dynamic parameters at the camshaft (the distribution mechanisms), when the cam was made to work normally. We can make the geometrical synthesis of the cam profile with the help of the cinematics of the mechanism. One uses as well the reduced speed, s’. The reduced velocity, s’, folded to 90 degrees, completes the triangle OAB. We can determine the coordinates of the point A from the tappet (1), and from the cam (2). The forces and the velocities at a cam with plate translated tappet can be seen in the figure 2. The driving force Fm, perpendicular on the r in A, is decomposed in two forces: the utile force Fu, which acts the tappet and the lost force Fa, who is a slipping force. The velocities take the same positions (system 4). The efficiency of the mechanism is determining with the relationship (5). We can make the geometro-kinematics synthesis of the cam profile with the help of the cinematics of the mechanism (see the Figure 3). Now, we can make the geometro-kinematics synthesis of the classic cam profile (system 7). The moments of inertia is determined with the relationships from the
system 8. The angular velocity w, and the angular acceleration, are determined with the presented relationships (system 9). For a classic cam and tappet mechanism (without valve) dynamic movement tappet is expressed by equation (10), who was presented and derived in Chapter 2 (equation 48), and now by canceling valve mass, will customize and reaching form below (10). Where x is the dynamic movement of the pusher, while s is its normal, kinematics movement. K is the spring constant of the system, and k is the spring constant of the tappet spring. It note, with x0 the tappet spring preload, with mT the mass of the tappet, with ω the angular rotation speed of the cam (or camshaft), where s’ is the first derivative in function
of of the tappet movement, s. Differentiating twice successively, the expression (10) in the angle , we obtain a reduced tappet speed (equation 11), and reduced tappet acceleration (12). The real and dynamic, tappet acceleration can be determined directly using the relation (13). The presented dynamic system has the advantage to has a normal functionality. The synthesis was made using the natural geometro-kinematics parameters (of cam mechanism).
Keywords: cam, cams, cam mechanisms, distribution mechanisms, camshaft, tappet.
1. Geometrical synthesis of the cam profile
We can make the geometrical synthesis of the cam profile with the help of the cinematics of the mechanism. One uses as well the reduced speed, s’. The reduced velocity, s’, folded to 90 degrees, completes the triangle OAB (see the Figure 1).
OB=r0+s; BA=s’; OA=r=rA; r2=rA2=(r0+s)2+s’2
It establishes a system fixed Cartesian, xOy = xfOyf, and a mobil Cartesian system, xOy = xmOym fixed with the cam.
From the lower position 0, the tappet, pushed by cam, uplifts to a general position, when
the cam rotates with the angle. The contact point A, go from Ai0 to A0 (on the cam), and to A (on
the tappet). The position angle of the point A from the tappet is f, and from the cam is m. We can determine the coordinates of the point A from the tappet (1), and from the cam (2).
fA
f
AT
fA
f
AT
rsryy
rsxx
sin
cos'
0
(1)
47
sin'cossincos
cossincossinsinsin
sincos'sincos
sinsincoscoscoscos
0
0
ssrxy
rrrryy
srsyx
rrrrxx
TT
fffmA
m
Ac
TT
fffmA
m
Ac
(2)
PA0
iA0A
O
Arr
fx
0r
f
0r0r
s
's
fy
mx
w
cam
tappet 0P
0
m
A
A
A
fm
r
s
r
sr
ssrr
'sin
cos
')(
0
22
0
2
fA
f
AT
fA
f
AT
rsryy
rsxx
sin
cos'
0
fffmA
m
Ac
fffmA
m
Ac
rrrryy
rrrrxx
cossincossinsinsin
sinsincoscoscoscos
sin'cossincos
sincos'sincos
0
0
ssrxyyy
srsyxxx
TT
m
Ac
TT
m
Ac
B
Fig. 1 Geometry of the cam with plate translated follower
48
One uses the relationships (3).
A
A
A
fm
r
s
r
sr
ssrr
'sin
cos
')(
0
22
0
2
(3)
Now, we shall see the forces, the powers and the efficiency.
2. The efficiency of the cam
The forces and the velocities at a cam with plate translated tappet can be seen in the figure 2.
PA0
iA0A
O
Arr
fx
0r
f
0r0r
s
's
fy
mx
w
TTu vFF ,
0
m
mm vF ,
vFFa ,
Fig. 2 Forces and velocities of the cam with plate translated follower
49
The driving force Fm, perpendicular on the r in A, is decomposed in two forces: the utile force Fu, which acts the tappet and the lost force Fa, who is a slipping force. The velocities take the same positions (system 4).
cos;sin
cos;sin
mamu
mamu
vvvv
FFFF (4)
The efficiency of the mechanism is determining with the relationship (5).
22 cossinsinsin
mm
mm
mm
uu
c
ui
vF
vF
vF
vF
P
P (5)
3. Geometro-kinematics synthesis of the cam profile
We can make the geometro-kinematics synthesis of the cam profile with the help of the cinematics of the mechanism (see the Figure 3).
A
0
iA0A
O
r
fx
0r
f
0r0r
s
ar
fy
mx
w
svT
0
m
mv
aa rv
B
Fig. 3 Geometro-kinematics synthesis of the cam with plate translated follower
50
We denote the distance BA with ra. We can write the relationships (6).
sr
ssrtg
ssrr
ssr
ssrr
sr
ssrrr
ssrrssrrssrr
dssrdrrdr
ds
sr
r
dr
ds
dt
drdt
ds
r
s
sr
r
aaa
aa
a
a
aaa
a
0
2
0
2
0
2
0
2
0
2
0
2
0
0
2
0
2
0
2
0
2
0
22
0
2
0
0
0
2
24
2sin
24cos
24
222
1
2
1
tan
(6)
Now, we can make the geometro-kinematics synthesis of the classic cam profile (system 7).
sin2cossincos
sincoscossincossinsinsin
sincos2sincos
sincossinsincoscoscoscos
sin
2cos
2
000
0
2
00
0
2
0
ssrsrrsr
xyrrrry
srssrsrr
yxrrrrx
srry
ssrrrx
a
TTfffmc
a
TTfffmc
fT
afT
(7)
For a law cos, the profile takes the below form, bean-shaped (see the Figure 4).
51
Fig. 4 The profile of cam (bean-shaped) to the cam with plate translated follower
This profile can be closed with an additional curve, and one obtains the form in the Figure 5.
52
-0.025
-0.02
-0.015
-0.01
-0.005
0
0.005
0.01
0.015
-0.015 -0.01 -0.005 0 0.005 0.01 0.015 0.02 0.025
yc
Fig. 5 The closed profile of cam to the cam with plate translated follower
53
4. The dynamics of the cam
The moments of inertia is determined with the relationships from the system 8.
2
0
*
2
0
2
0
2
0
*
2
0
2
0
*
0
2
0
0
22
0
min
2
0
*
22
0
2
0
22
0
2
0
*
22
0
2
0
2
2
1
2
1
2
1
2
1
2
12
2
1
2
1
1cos0''
20'
0'
'''2'2'2'
'2
;0',00;2
1
2
1
'2
;2
1'24
2
1
';242
1
2
1
hrMJ
hrMhMhrMrMJ
hMhrMrMJ
IIsmsMrM
IhMhrMJhss
J
ssmssMsrMJ
smsMsrMMaxJJ
sswhenJwithJrMJ
smsMsrMJwith
JrMsmssrrMJ
smJssrrMrMJ
Cm
Ccccm
cccm
Tcc
ccM
Tcc
TccMMax
Mcm
Tcc
cTc
TTccc
(8)
The angular velocity w, and the angular acceleration, are determined with the presented relationships (system 9).
'''2'2'2'
'242
1
2
1
2
1
;
0
*'
22
0
2
0
*
2
0
*
*
*'2
*
*2
*
*2
ssmssMsrMJJ
smssrrMJ
hrMJ
J
J
J
J
J
J
Tcc
Tc
Cm
mm
mm
w
wwww
(9)
For a classic cam and tappet mechanism (without valve) dynamic movement tappet is expressed by equation (10), who was presented and derived in Chapter 2 (equation 48), and now by canceling valve mass, will customize and reaching form below (10).
54
kK
xkskK
skKxksKkksmkKsx T
02
0
2222
)(2
)(2)2(')( w (10)
Where x is the dynamic movement of the pusher, while s is its normal, kinematics movement. K is the spring constant of the system, and k is the spring constant of the tappet spring. It note, with x0 the tappet spring preload, with mT the mass of the tappet, with ω the angular
rotation speed of the cam (or camshaft), where s’ is the first derivative in function of of the tappet
movement, s. Differentiating twice successively, the expression (10) in the angle , we obtain a reduced tappet speed (equation 11), and reduced tappet acceleration (12).
2
02
0
0
22
0
2222
2
''
'
'2'22'''2
)(2)2(')(
kK
kxskK
Msx
sNkK
kxs
skKkxsskKkssmkKM
skKxksKkksmkKN
T
T
w
w
(11)
3
02
00
0
22
22
0
0
22
0
2222
2
'2''
''''
''2'''22
''''''2
'
'2'22'''2
)(2)2(')(
kK
kxskK
sMkK
kxssN
kK
kxsO
sx
skKxksssKkk
sssmkKO
sNkK
kxs
skKkxsskKkssmkKM
skKxksKkksmkKN
T
T
T
w
w
w
(12)
The real and dynamic, tappet acceleration can be determined directly using the relation (13).
w ''' 2 xxx (13)
55
For a law cos (14) we obtain the dynamic diagram from the Figure 6.
uu
r
uu
r
uu
r
u
hs
has
hvs
hhs
sin2
'''
cos2
''
sin2
'
cos22
3
3
2
2 (14)
Fig. 6 The dynamic diagram of the tappet acceleration from a cos profile of cam used to the cam with plate
translated follower; r0=13 [mm], Mc=200 [g], mT=100 [g], u=c=/2, h=6 [mm], nm=10000 [rpm], x0=90 [mm], k=40 [kN/m], K=5000 [kN/m].
5. Conclusions
The presented dynamic system has the advantage to has a normal functionality. The synthesis was made using the natural geometro-kinematics parameters (of cam mechanism).
56
CHAPTER VII DYNAMIC SYNTHESIS OF THE ROTARY CAM AND TRANSLATED
TAPPET WITH ROLL
Abstract: This chapter presents an original methods to determine the dynamic parameters at the camshaft (the distribution mechanisms). We determine initially the mass moment of inertia (mechanical) of the mechanism, reduced to the element of rotation, ie at cam (basically using kinetic energy conservation, the system 1). The rotary cam with translated follower with roll (Figure 1), is synthesized dynamic. We considered the law of motion of the tappet classic version already used the cosine law (both ascending and
descending). The angular velocity is a function of the cam position () but also its rotation speed (2). Where ωm is the nominal angular velocity of cam and express at the distribution mechanisms based on the motor shaft speed (3). We start the simulation with a classical law of motion, namely the cosine law. To climb cosine law system is expressed by relations (4). With the relation (5) is expressed the first derivative of the reduced mechanical moment of inertia. It is necessary to determine the angular acceleration (6). Relations (2) and (6) a general nature and is basically two original equations of motion crucial for mechanical mechanisms. For a rotary cam and translated tappet with roll mechanism (without valve), dynamic movement tappet is expressed by equation (7), who was presented and derived in Chapter 2 (equation 48), and now by canceling valve mass, will customize and reaching form below (7). Where x is the dynamic movement of the pusher, while s is its normal, kinematics movement. K is the spring constant of the system, and k is the spring constant of the tappet spring. It note, with x0 the tappet spring preload, with mT the mass of the tappet, with ω the angular rotation speed of the cam (or camshaft), where s’ is the first derivative in
function of of the tappet movement, s. Differentiating twice successively, the expression (7) in the angle , we obtain a reduced tappet speed (equation 8), and reduced tappet acceleration (9). Further the acceleration of the tappet can be determined directly real (dynamic) using the relation (10). For a good work one proposes to make a new geometro-kinematics synthesis of the cam profile, using some new relationships (16).
Keywords: cam, cams, cam mechanisms, distribution mechanisms, camshaft, tappet.
The rotary cam with translated follower with roll (Figure 1), is synthesized dynamic by the next relationships.
0A
A
B
A-
B
B0
A0
A
O
X’
e
s0
rb
r0
rA
rB
s
n
C
rb
x
y
s’
Fig. 1 The rotary cam with translated follower with roll
57
First, one determines the mass moment of inertia (mechanical) of the mechanism, reduced to the element of rotation, ie cam (basically using kinetic energy conservation, system 1).
2
22
0
2
0
2
2
00
2
0
2*
2
0
22
2
0
2
0
2
0
2
0
2
00
22
0
*
22
0
22
0
2
02
0
222
22
0
00
22
0
2
0
222
0
2
0
222
0
222
0
2222222
2
''
'
2
122
2
1
8
22
22
2
1
16
1
4
1
2
1
'
'2
'
2
'2
'
'2
cossin2
cos2cos
sin2sin
2
1
smesss
sseserM
sMssMrrrrMJ
hm
ehh
s
hse
he
rM
hMhsMrrrrMJ
esss
eser
esss
ssrssrer
esss
ssssr
esss
eserssrer
sserssrer
ssrrss
rereyxrR
RMJ
Tbc
ccbbc
Tbc
ccbbcm
b
bbA
b
bbA
bbA
bb
bbAAA
ccama
(1)
58
We considered the law of motion of the tappet classic version already used the cosine law (both ascending and descending).
The angular velocity is a function of the cam position () but also its rotation speed (2). Where ωm is the nominal angular velocity of cam and express at the distribution mechanisms based on the motor shaft speed (3).
2
*
*2
mm
J
Jww (2)
60260
2
6022
nnn motorccm
w (3)
We start the simulation with a classical law of motion, namely the cosine law. To climb cosine law system is expressed by relations (4).
uu
r
uu
r
uu
r
u
hs
has
hvs
hhs
sin2
'''
cos2
''
sin2
'
cos22
3
3
2
2 (4)
Where takes values from 0 to u.
Jmax occurs for =u/2.
With the relation (5) is expressed the first derivative of the reduced mechanical moment of inertia. It is necessary to determine the angular acceleration (6).
2/322
0
0
2
0
2
2/322
0
22
00
0
*'
'
'''''
'
''2''
'''2''
esss
sesssssseserM
esss
essssssserM
ssmssMssMJ
bc
bc
Tcc
(5)
59
Differentiating the formula (2), against time, is obtained the angular acceleration expression (6).
*
*'2
2 J
J
w (6)
Relations (2) and (6) a general nature and is basically two original equations of motion crucial for mechanical mechanisms.
For a rotary cam and translated tappet with roll mechanism (without valve), dynamic movement tappet is expressed by equation (7), who was presented and derived in Chapter 2 (equation 48), and now by canceling valve mass, will customize and reaching form below (7).
kK
xkskK
skKxksKkksmkKsx T
02
0
2222
)(2
)(2)2(')( w
(7)
Where x is the dynamic movement of the pusher, while s is its normal, kinematics movement. K is the spring constant of the system, and k is the spring constant of the tappet spring.
It note, with x0 the tappet spring preload, with mT the mass of the tappet, with ω the angular
rotation speed of the cam (or camshaft), where s’ is the first derivative in function of of the tappet
movement, s. Differentiating twice successively, the expression (7) in the angle , we obtain a reduced tappet speed (equation 8), and reduced tappet acceleration (9).
2
02
0
0
22
0
2222
2
''
'
'2'22'''2
)(2)2(')(
kK
kxskK
Msx
sNkK
kxs
skKkxsskKkssmkKM
skKxksKkksmkKN
T
T
w
w
(8)
60
3
02
00
0
22
22
0
0
22
0
2222
2
'2''
''''
''2'''22
''''''2
'
'2'22'''2
)(2)2(')(
kK
kxskK
sMkK
kxssN
kK
kxsO
sx
skKxksssKkk
sssmkKO
sNkK
kxs
skKkxsskKkssmkKM
skKxksKkksmkKN
T
T
T
w
w
w
(9)
Further the acceleration of the tappet can be determined directly real (dynamic) using the relation (10).
w ''' 2 xxx (10)
Dynamic synthesis
Give the following parameters:
r0=0.013 [m]; rb=0.005 [m]; h=0.008 [m]; e=0.01 [m]; x0=0.03 [m]; u=/2; c=/2; K=5000000 [N/m]; k=20000 [N/m]; mT=0.1 [kg]; Mc=0.2 [kg]; nmotor=5500 [rot/min].
61
To sum up dynamically based on a computer program, you can vary the input data until the corresponding acceleration is obtained (see Figure 2). It then summarizes the corresponding cam profile (Figure 3) using the relations (11).
Fig. 2 Dynamic diagram to the rotary cam with translated follower with roll
coscossinsin
sincoscossin
cossin
sincos
cos
sin
0
0
0
bbC
bbC
TTC
TTC
bT
bT
rssrey
rssrex
yxy
yxx
rssy
rex
(11)
62
Fig. 3 The cam profile to the rotary cam with translated follower with roll
rb=0.003 [m]; e=0.003 [m]; h=0.006 [m]; r0=0.013 [m]; 0=/2 [rad];
The geometry of the rotary cam and the translated follower with roll
Now, we shall see the geometry of a rotary cam with translated follower with roll (Figure 4). The cam rotation sense is positive (trigonometric).
We can make the geometrical synthesis of the cam profile with the help of the cinematics of the mechanism. One uses as well the reduced speed, s’.
OA=r=rA; r2=rA2
It establishes a system fixed Cartesian, xOy = xfOyf, and a mobil Cartesian system, xOy = xmOym fixed with the cam.
From the lower position 0, the tappet, pushed by cam, uplifts to a general position, when
the cam rotates with the angle. The contact point A, go from Ai0 to A0 (on the cam), and to A (on
the tappet). The position angle of the point A from the tappet is f, and from the cam is m. We can determine the coordinates of the point A from the tappet (12), and from the cam (13).
63
f
m
A
0
iA
0AO
fx
fy
mx
0B
B
e
s
0s
w
r
0r
0r
br
br
Br
0
Fig. 4 The geometry of the rotary cam with translated follower with roll
64
ffAb
f
AT
ffAb
f
AT
rrrssyy
rrrexx
sinsincos
coscossin
0
(12)
sinsincoscossincos
cossincossinsinsin
sincoscossinsincos
sinsincoscoscoscos
0
0
bbTT
fffmA
m
Ac
bbTT
fffmA
m
Ac
rerssxy
rrrryy
rssreyx
rrrrxx
(13)
One uses and the next relationships (where the pressure angle was obtained with the classic Antonescu P. method):
ss
estg
esss
es
esss
ss
errs b
0
22
0
22
0
0
22
00
'
'
'sin
'cos
(14)
Determining the forces, the velocities and the efficiency (15)
The driving force Fm, perpendicular on r in A, is divided in two components: Fn, the normal force, and Fa, a force of slipping. Fn is divided, as well, in two components: FT is the transmitted (the utile) force, and FR is a radial force which bend the tappet (see 15, and the Figure 5).
cossin2
cossin
;;2
cos;22
;
coscoscoscoscoscoscoscos
coscoscos
coscoscos
cos
cos
0
2
0
22
2
0
222
2
0
2222
222
sserssrerr
rssreyxr
sserrr
rrrAAA
vF
vF
vF
vF
P
P
vvv
FFF
vv
FF
bbA
bbAfAf
B
b
Bb
mm
mm
mm
TT
c
ui
mnT
mnT
mn
mn
(15)
65
f
m
A
0
iA
0AO
fx
fy
mx
0B
B
e
s
0s
w
r
0r
0r
br
br
Br
0
TF
RF
nF
nF
mF
aF
Fig. 5 Forces and velocities of the rotary cam with translated follower with roll
66
Geometro-kinematics synthesis
For a good work one proposes to make a new geometro-kinematics synthesis of the cam profile, using some new relationships (16).
ffAb
f
AT
ffAb
f
AT
rrrssyy
rrrexx
sinsincos
coscossin
0
(12)
sinsincoscossincos
cossincossinsinsin
sincoscossinsincos
sinsincoscoscoscos
0
0
bbTT
fffmA
m
Ac
bbTT
fffmA
m
Ac
rerssxy
rrrryy
rssreyx
rrrrxx
(13)
One uses and the next relationships (where the pressure angle was obtained with the new Petrescu F. method):
][2
'2'4'4arccossinsin
][2
'2'4'4arccos
cosarccos
][2
'2'4'4cos
22
0
22
00
2
0
22
0
22
00
2
0
22
0
22
00
2
0
22
00
ess
sesesssssss
ess
sesesssssss
ess
sesesssssss
errs b
(16)
67
The new profile can be seen in the Figure 6.
Fig. 6 The new cam profile to the rotary cam with translated follower with roll
rb=0.003 [m]; e=0.003 [m]; h=0.006 [m]; r0=0.013 [m]; 0=/2 [rad];
68
Demonstration (explication)
][2
'2'4'4cos
][2
'4'2'2cos
0'cos'2cos
cos'2cos'coscos
cos'cos1cos
coscossin'cossin
cos
'
cossincossincossinsin
sincos
'sin
cos
'
sinsin
sincos
'
sin2
cos2
coscos
22;;
2
coscos'coscos
22
0
22
00
2
0
22
0
22
0
222
0
2
02
222
0
422
0
242242
0
22
0
22
0
2
00
0
ess
sesesssssss
ess
essssessesss
ssessess
seesssss
esss
esssr
ess
r
s
r
e
r
ssBBB
Br
sB
r
r
r
s
Br
rA
Ar
s
AAA
AA
rsrs
BB
BB
BA
B
A
A
B
A
AA
w
(17)
69
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223.-T3. TEMPEA I., BALESCU C., ADIR G., Mecanism de presare destinat mecanizãrii operatiei de formare în rame (pãrtile I si II). In al VII-lea Simpozion national de roboti industriali si mecanisme spatiale. Vol. 3., Bucuresti, 1987.
224.-T4. TEMPEA I., GRADU M., Sinteza camei de translatie cu tachet cu rolã, cu ajutorul functiilor spline. In lucrãrile simpozionului de R.I., Timisoara, 1992.
225.-T5. TUTUNARU D., Mecanisme plane rectiliniare si inversoare. Editura tehnicã, Bucuresti, 1969.
226.-T6. TORAZZA G., A variable lift and event control device piston engine valve operation. In FISITA XIV Congres,Paper II / 10, London, 1972.
227.-T7. TESAR D., MATTHEW G.K., The design of modelled cam sistems. In Cams and cam mechanisms, 1974.
228.-T8. TERME D., Besondere Merkmalebeider Nutzung des Pressungwinkels fur kurvengetriebeanalyse und-Synthese. In SYROM’85,Vol. III-2, pp. 489-504, Bucuresti, iulie 1985.
229.-T9. TEMPEA I., DUGĂEŞESCU I., NEACŞA M., Mecanisme. Noţiuni teoretice şi teme de proiect rezolvate, Ed. Printech, ISBN (10) 973-718-560-9, 2006.
230.-T10. D. Taraza, N.A. Henein, W. Bryzik, "The Frequency Analysis of the Crankshaft's Speed Variation: A reliable Tool for Diesel Engine Diagnosis," ASME Journal for Gas Turbines and Power 123(2), 428-432, 2001 231.-T11. D. Taraza, "Accuracy Limits of IMEP Determination from Crankshaft Speed Measurements," SAE Transactions, Journal of Engines 111, 689-697, 2002.
232.-T12. D. Taraza, "Statistical Correlation Between the Crankshaft's Speed Variation and Engine Performance, Part I: Theoretical Model," ASME Journal of Engineering for Gas Turbines and Power 125(3), 791-796, 2003.
233.-T13. D. Taraza, "Statistical Correlation Between the Crankshaft's Speed Variation and Engine Performance, Part II: Detection of Deficient Cylinders and MIP Calculation," ASME journal of Engineering for Gas Turbines and Power 125(3), 797-803, 2003.
234.-U1. ULF A., WILLIAM S., A Simple Procedure for Modifying High-Speed Cam Profiles for Vibration Reduction, Journal of Mechanical Design - November 2004 - Volume 126, Issue 6, pp. 1105-1108.
235.-V1. VOINEA R., VOICULESCU D., CEAUSU V., Mecanica. E.D.P., Bucuresti, 1975.
236.-V2. VOINEA R., ATANASIU M., Metode analitice noi în teoria mecanismelor. Editura tehnicã, Bucuresti, 1964.
237.-V3. Van de Straete, H.J., De Schutter, J., Hybrid cam mechanisms, Mechatronics, IEEE/ASME Transactions on Volume 1, Issue 4, Dec. 1996 Page(s):284 - 289
238.-W1. WIEDERRICH J.L., ROTH B., Design of low vibration cam profiles. In Cams and cam mechanisms, Edited by J. REES JONES, MEP, London and Birmingham, Alabama, 1974.
239.-W2. WIEDERRICH J.L., ROTH B., Dynamic Synthesis of Cams Using Finite Trigonometric Series, Trans. ASME, 1974.
240.-Y1. YOUNG V.C., Considerations în valve gear design. Trans. SAE, 1, 1947, pp. 359-365.
241.-Z1. ZHANG J.L., LI Z., Research on the dynamics of a RSCR spatial mechanisms considering bearing clearances. In al VI-lea SYROM, Vol. II, Bucuresti, iunie 1993.
Annex
Profile of rotary cam and translated lift with roller; Law of motion ex:
-10
-5
0
5
10
15
-20 -10 0 10
u= 85[grad]
c= 85[grad]
r0= 3[mm]
rb = 15[mm]
e= 0[mm]
hT= 12[mm]
Law
y=(ex -e
-x)/a
w
Down profile
Up profile
Profile of rotary cam and translated lift with roll; Law of motion ex;
Good efficiency =25%; Name:Name: ““Elax001Elax001”” nncamcam=n/2=n/2
d0=6 [mm]
d0=6 [mm], h3 [mm], rb=15 [mm]; d0=12 [mm], h6 [mm], rb=30 [mm];
d0=24 [mm], h12 [mm], rb=60 [mm].
76
-25000
-20000
-15000
-10000
-5000
0
5000
10000
0 50 100 150 200
a[m/s2]
312,85s*k[mm] k=
n=5500[rot/min]
ju=85 [grad]
k=50 [N/mm]
r0=3 [mm]
x0=80 [mm]
hs=12 [mm]
hT=12 [mm]
i=1;h=51.1%
rb=15 [mm]
e=0 [mm]
y=(ex-e
-x)/a
a=2.35040238
amax=4130smax=10.55
amin= -21300
Dynamic analysis to the rotary cam and translated follower with roll;
Law of motion ex; Name:Name: ““Elax001Elax001””
PProposed crankshaft rotation speed n=5500 [r/m]; roposed crankshaft rotation speed n=5500 [r/m]; nncamcam=n/2=n/2
-initial spring deflection x0=80 [mm]
-elastic constant of spring k=50-80 [N/mm]
-roll radius of the follower rb=15 [mm] at a r0=3 [mm].
Profile of rotary cam and translated lift with roller; Law of motion ex:
-10
-5
0
5
10
15
-20 -10 0 10
u= 85[grad]
c= 85[grad]
r0= 3[mm]
rb = 15[mm]
e= 0[mm]
hT= 12[mm]
Law
y=(ex -e
-x)/a
w
Down profile
Up profile
Profile of rotary cam and translated lift with roll; Law of motion ex;
Best efficiency =51%; Name:Name: ““2Elax0012Elax001”” ![![nncamcam=n/4=n/4]!]!
d0=6 [mm]
d0=6 [mm], h3 [mm], rb=15 [mm];
d0=12 [mm], h6 [mm], rb=30 [mm];
d0=24 [mm], h12 [mm], rb=60 [mm].
y=(ex-e-x)/a
a=2.35040238
77
-10
-5
0
5
10
-20 -10 0 10
u= 55[grad]
c= 55[grad]
r0= 4[mm]
rb = 5[mm]
e= 0[mm]
hT= 12[mm]
Law LOG-0
a=1
profil dublat
w
Profile of rotary cam and translated lift with roll; Law of motion Log Nat;
Good efficiency =25-33%; Name:Name: ““2LogNat0012LogNat001”” ![![nncamcam=n/4=n/4]!]!
Up profile
Down profile
d0=8 [mm], h3,5 [mm], rb=5 [mm];
d0=16 [mm], h7 [mm], rb=10 [mm];
d0=24 [mm], h10,5 [mm], rb=15 [mm].
y=(ln(x+a)-ln(a))/
(ln(a+1)-ln(a))
-10000
-8000
-6000
-4000
-2000
0
2000
4000
6000
0 50 100 150
a[m/s2]
288,48s*k[mm] k=
n=5500[rot/min]
ju=55 [grad]
k=30 [N/mm]
r0=4 [mm]
x0=80 [mm]
hs=12 [mm]
hT=12 [mm]
i=1;h=32.3%
rb=5 [mm]
e=0 [mm]
legea:Log-0
a=1
amax=4040smax=11.21
amin= -8400
y=(ln(x+a)-ln(a))/(ln(a+1)-ln(a))
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion Log Nat;
Good efficiency =25-33%; Name:Name: ““2LogNat0012LogNat001”” ![![nncamcam=n/4=n/4]!]!
-initial spring deflection x0=80 [mm]
-elastic constant of spring k=30-60 [N/mm]
-roll radius of the follower rb=5 [mm] at a r0=4 [mm].
78
-10
-5
0
5
10
15
20
-20 -15 -10 -5 0 5 10
w
Profile of rotary cam and translated lift with roll; Law of motion Power;
Better efficiency =40-47%; Name:Name: ““2Power0012Power001”” ![![nncamcam=n/4=n/4]!]!
d0=8 [mm], h5 [mm], rb=20 [mm];
d0=16 [mm], h10 [mm], rb=40 [mm];
d0=24 [mm], h15 [mm], rb=60 [mm].
Up profile
Down profile
-10
-5
0
5
10
15
20
-20 -15 -10 -5 0 5 10
yC [mm]
ju= 85[grad]
jc= 85[grad]
r0= 4[mm]
rb = 20[mm]
e= 0[mm]
hT= 12[mm]
The Law
Power
y=2x-1
w
-25000
-20000
-15000
-10000
-5000
0
5000
10000
0 50 100 150 200
a[m/s2]
330,54s*k[mm] k=
n=5500[rot/min]
ju=85 [grad]
k=100 [N/mm]
r0=4 [mm]
x0=80 [mm]
hs=12 [mm]
hT=12 [mm]
i=1;h=47.2%
rb=20 [mm]
e=0 [mm]
law:Power-0
y=2x-1
amax=4000smax=9.55
amin= -22000
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion Power;
Better efficiency =40-47%; Name:Name: ““2Power0012Power001”” ![![nncamcam=n/4=n/4]!]!
-initial spring deflection x0=80 [mm]
-elastic constant of spring k=100 [N/mm]
-roll radius of the follower rb=20 [mm] at a r0=4 [mm].
79
-15
-10
-5
0
5
10
15
-15 -10 -5 0 5 10 15
u= 75[grad]
c= 75[grad]
r0= 6[mm]
rb = 14[mm]
e= 0[mm]
hT= 8[mm]
Law ATAN
w
Up profile
Down profile
y=arctg(x/a)
a=0.642
Profile of rotary cam and translated lift with roll; Law of motion Atan;
Good efficiency =30-40%; Name:Name: ““2Atan0012Atan001”” ![![nncamcam=n/4=n/4]!]!
d0=12 [mm], h3 [mm], rb=14 [mm];
d0=24 [mm], h6 [mm], rb=28 [mm];
d0=36 [mm], h9 [mm], rb=42 [mm].
-10000
-8000
-6000
-4000
-2000
0
2000
4000
6000
8000
10000
12000
0 50 100 150 200
a[m/s2]
1000,69s*k[mm] k=
n=5500[rot/min]
ju=75 [grad]
k=30 [N/mm]
r0=6 [mm]
x0=80 [mm]
hs=8 [mm]
hT=8 [mm]
i=1;h=28.4%
rb=14 [mm]
e=0 [mm]
law: ATAN-1
y=arctg(x/a)
a=0.642
amax=9229
smax=7.37
amin= -7753
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion Atan;
Good efficiency =30-40%; Name:Name: ““2Atan0012Atan001”” ![![nncamcam=n/4=n/4]!]!
-initial spring deflection x0=80 [mm]
-elastic constant of spring k=30 [N/mm]
-roll radius of the follower rb=14 [mm] at a r0=6 [mm].
80
-20
-15
-10
-5
0
5
10
15
20
-30 -20 -10 0 10 20
u= 65[grad]
c= 65[grad]
r0= 13[mm]
rb = 12[mm]
e= 0[mm]
hT= 8[mm]
Law 2C4P1-001
y=2x-x2
w
Profile of rotary cam and translated lift with roll; Law of motion C4P1-001;
Good efficiency =20-35%; Name:Name: ““2C4P12C4P1--001001”” ![![nncamcam=n/4=n/4]!]!
d0=26 [mm], h6 [mm], rb=12 [mm];
d0=52 [mm], h12 [mm], rb=24 [mm];
Up profile
Down profile
-2000
0
2000
4000
6000
8000
10000
12000
0 50 100 150 a[m/s2]
1054,98s*k[mm] k=
n=5500[rot/min]
u=65 [grad]
k=50 [N/mm]
r0=13 [mm]
x0=20 [mm]
hs=8 [mm]
hT=8 [mm]
i=1;=16.7%
rb=12 [mm]
e=0 [mm]
law: 2C4P1-001
y=2x-x2
amax=10100
smax=7.66
amin= -260
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion C4P1-001;
Good efficiency =20-35%; Name:Name: ““2C4P12C4P1--001001”” ![![nncamcam=n/4=n/4]!]!
-initial spring deflection x0=20 [mm]
-elastic constant of spring k=50 [N/mm]
-roll radius of the follower rb=12 [mm] at a r0=13 [mm].
81
-20
-15
-10
-5
0
5
10
15
20
-30 -20 -10 0 10 20
u= 45[grad]
c= 45[grad]
r0= 13[mm]
rb = 12[mm]
e= 0[mm]
hT= 8[mm]
Law 4C4P1-001
y=2x-x2
w
Profile of rotary cam and translated lift with roll; Law of motion 4C4P1-001;
Best efficiency =40-70%; Name:Name: ““4C4P14C4P1--001001”” ![![nncamcam=n/8=n/8]!]!
d0=26 [mm], h3 [mm], rb=12 [mm];
d0=52 [mm], h6 [mm], rb=24 [mm];
d0=78 [mm], h9 [mm], rb=36 [mm].
Up profile
Down profile
-1000
0
1000
2000
3000
4000
5000
6000
7000
0 20 40 60 80 100 a[m/s2]
638,88s*k[mm] k=
n=5500[rot/min]
u=45 [grad]
k=35 [N/mm]
r0=13 [mm]
x0=20 [mm]
hs=8 [mm]
hT=8 [mm]
i=1;=23.5%
rb=12 [mm]
e=0 [mm]
law: 4C4P1-001
y=2x-x2
amax=6200
smax=7.76
amin= -140
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion C4P1-001;
Best efficiency =40-70%; Name:Name: ““4C4P14C4P1--001001”” ![![nncamcam=n/8=n/8]!]!
-initial spring deflection x0=20 [mm]
-elastic constant of spring k=35 [N/mm]
-roll radius of the follower rb=12 [mm] at a r0=13 [mm].
82
-15
-10
-5
0
5
10
15
20
-20 -10 0 10 20
ju= 45[grad]
jc= 45[grad]
r0= 13[mm]
rb = 3[mm]
e= 0[mm]
hT= 6[mm]
Law SIN
w
y=x-sin(2x)/(2)
Profile of rotary cam and translated lift with roll; Law of motion SIN-001;
Good efficiency =15-30%; Name:Name: ““2SIN2SIN--001001”” ![![nncamcam=n/4=n/4]!]!
d0=26 [mm], h6 [mm], rb=3 [mm];
d0=52 [mm], h12 [mm], rb=6 [mm].
Up profile
Down profile
-3000
-2000
-1000
0
1000
2000
3000
0 20 40 60 80 100
a[m/s2]
363,66s*k[mm] k=
n=5500[rot/min]
ju=45 [grad]
k=15 [N/mm]
r0=13 [mm]
x0=30 [mm]
hs=6 [mm]
hT=6 [mm]
i=1;h=15.3%
rb=3 [mm]
e=0 [mm]
law: sin
y=x-sin(2px)/(2p)
amax=2667
smax=5.87
amin= -1890
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion SIN-001;
Good efficiency =15-30%; Name:Name: ““2SIN2SIN--001001”” ![![nncamcam=n/4=n/4]!]!
-initial spring deflection x0=30 [mm]
-elastic constant of spring k=15 [N/mm]
-roll radius of the follower rb=3 [mm] at a r0=13 [mm].
83
-25
-20
-15
-10
-5
0
5
10
15
20
25
-30 -20 -10 0 10 20 30
ju=22,5[grad]
jc=22,5[grad]
r0= 20[mm]
rb = 1[mm]
e= 0[mm]
hT= 6[mm]
Law SIN
y=x-sin(2x)/(2 )
Profile of rotary cam and translated lift with roll; Law of motion SIN-001;
Better efficiency =15-60%; Name:Name: ““4SIN4SIN--001001”” ![![nncamcam=n/8=n/8]!]!
d0=40 [mm], h6 [mm], rb=1 [mm];
d0=60 [mm], h9 [mm], rb=1,5 [mm].
Up profile
Down profile
w
-3000
-2000
-1000
0
1000
2000
3000
4000
0 10 20 30 40 50
a[m/s2]
442,41s*k[mm] k=
n=5500[rot/min]
ju=22,5 [grad]
k=5 [N/mm]
r0=20 [mm]
x0=100 [mm]
hs=6 [mm]
hT=6 [mm]
i=1;h=15.7%
rb=1 [mm]
e=0 [mm]
law: sin
y=x-sin(2px)/(2p)
amax=3250
smax=5.87
amin= -2030
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion SIN-001;
Better efficiency =15-60%; Name:Name: ““4SIN4SIN--001001”” ![![nncamcam=n/8=n/8]!]!
-initial spring deflection x0=100 [mm]
-elastic constant of spring k=5 [N/mm]
-roll radius of the follower rb=1 [mm] at a r0=20 [mm].
84
d0=40 [mm], h6 [mm], rb=4 [mm];
d0=60 [mm], h9 [mm], rb=6 [mm].
-25
-20
-15
-10
-5
0
5
10
15
20
25
-30 -20 -10 0 10 20 30
ju=22,5[grad]
jc=22,5[grad]
r0= 20[mm]
rb = 4[mm]
e= 0[mm]
hT= 6[mm]
Law SIN
y=x-sin(2x)/(2 )
Profile of rotary cam and translated lift with roll; Law of motion SIN-002;
Better efficiency =17-70%; Name:Name: ““4SIN4SIN--002002”” ![![nncamcam=n/8=n/8]!]!
Down profile
Up profile
w
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion SIN-002;
Better efficiency =17-70%; Name:Name: ““4SIN4SIN--002002”” ![![nncamcam=n/8=n/8]!]!
-3000
-2000
-1000
0
1000
2000
3000
4000
5000
0 10 20 30 40 50
a[m/s2]
546,62s*k[mm] k=
n=5500[rot/min]
ju=22,5 [grad]
k=5 [N/mm]
r0=20 [mm]
x0=100 [mm]
hs=6 [mm]
hT=6 [mm]
i=1;h=17.95%
rb=4 [mm]
e=0 [mm]
law: sin
y=x-sin(2px)/(2p)
amax=4000
smax=5.87
amin= -2000
-initial spring deflection x0=100 [mm]
-elastic constant of spring k=5 [N/mm]
-roll radius of the follower rb=4 [mm] at a r0=20 [mm].
85
-25
-20
-15
-10
-5
0
5
10
15
20
25
-30 -20 -10 0 10 20 30
ju=11,25[grad]
jc=11,25[grad]
r0= 20[mm]
rb = 1[mm]
e= 0[mm]
hT= 6[mm]
Law SIN
y=x-sin(2x)/(2 )
Profile of rotary cam and translated lift with roll; Law of motion SIN-001;
Better efficiency =14-80%; Name:Name: ““8SIN8SIN--001001”” ![![nncamcam=n/16=n/16]!]!
d0=40 [mm], h6 [mm], rb=1 [mm];
d0=60 [mm], h9 [mm], rb=1.5 [mm];
d0=80 [mm], h12 [mm], rb=2 [mm].
w
Up profile
Down profile
-4000
-2000
0
2000
4000
6000
8000
10000
12000
14000
16000
0 5 10 15 20 25a[m/s2]
1977,24s*k[mm] k=
n=5500[rot/min]
ju=11,25 [grad]
k=30 [N/mm]
r0=20 [mm]
x0=85 [mm]
hs=6 [mm]
hT=6 [mm]
i=1;h=14.0%
rb=1 [mm]
e=0 [mm]
law: sin
y=x-sin(2px)/(2p)
amax=13300
smax=5.39
amin= -1700
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion SIN-001;
Better efficiency =14-80%; Name:Name: ““8SIN8SIN--001001”” ![![nncamcam=n/16=n/16]!]!
-initial spring deflection x0=85 [mm]
-elastic constant of spring k=30 [N/mm]
-roll radius of the follower rb=1 [mm] at a r0=20 [mm].
86
-40
-30
-20
-10
0
10
20
30
40
-60 -40 -20 0 20 40
ju= 15[grad]
jc= 15[grad]
r0= 30[mm]
rb = 3[mm]
e= 0[mm]
hT=8[mm]
Law COS
y=.5-.5cos(x)
w
Profile of rotary cam and translated lift with roll; Law of motion COS-001;
Better efficiency =21-60%; Name:Name: ““6COS6COS--001001”” ![![nncamcam=n/12=n/12]!]!
d0=60 [mm], h8 [mm], rb=3 [mm];
d0=90 [mm], h12 [mm], rb=4.5 [mm].
Down profile
Up profile
w
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion COS-001;
Better efficiency =21-60%; Name:Name: ““6COS6COS--001001”” ![![nncamcam=n/12=n/12]!]!
-initial spring deflection x0=100 [mm]
-elastic constant of spring k=37 [N/mm]
-roll radius of the follower rb=3 [mm] at a r0=30 [mm].
-2000
-1000
0
1000
2000
3000
4000
5000
6000
0 10 20 30 40a[m/s2]
626,79s*k[mm] k=
n=5500[rot/min]
ju=15 [grad]
k=37 [N/mm]
r0=30 [mm]
x0=100 [mm]
hs=8 [mm]
hT=8 [mm]
i=1;h=21.0%
rb=3 [mm]
e=0 [mm]
law: cos
y=.5-.5cos(px)
amax=5600
smax=7.11
amin= -1400
87
Profile of rotary cam and translated lift with roll; Law of motion COS-001;
Best efficiency =19-80%; Name:Name: ““12COS12COS--001001”” ![![nncamcam=n/24=n/24]!]!
-60
-40
-20
0
20
40
60
-80 -60 -40 -20 0 20 40 60
ju= 7,5[grad]
jc= 7,5[grad]
r0= 50[mm]
rb = 2[mm]
e= 0[mm]
hT=8[mm]
Law COS
y=.5-.5cos(x)
w
d0=50 [mm], h4 [mm], rb=1 [mm].
d0=100 [mm], h8 [mm], rb=2 [mm];
d0=150 [mm], h12 [mm], rb=3 [mm].
Down profile
Up profile
w
It would be desirable rounding the peaks!
Dynamic analysis to the rotary cam and translated lift with roll; Law of motion COS-001;
Best efficiency =19-80%; Name:Name: ““12COS12COS--001001”” ![![nncamcam=n/24=n/24]!]!
-initial spring deflection x0=100 [mm]
-elastic constant of spring k=50 [N/mm]
-roll radius of the follower rb=2 [mm] at a r0=50 [mm].
-2000
-1000
0
1000
2000
3000
4000
5000
0 5 10 15 20
a[m/s2]
474,38s*k[mm] k=
n=5500[rot/min]
ju=7,5 [grad]
k=50 [N/mm]
r0=50 [mm]
x0=100 [mm]
hs=8 [mm]
hT=8 [mm]
i=1;h=19.19%
rb=2 [mm]
e=0 [mm]
law: cos
y=.5-.5cos(px)
amax=4060
smax=6.85
amin= -1400
!All these matters are copyrighted!
88